Chapter 11, Problem 18CT

To determine

Find the distance between given point and equation of plane.

Answer to Problem 18CT

The distance between the points $\left(2,–1,6\right)$ and the plane $3x–2y+z=6$ is $\frac{11\sqrt{14}}{14}$ .

Explanation of Solution

Given:

The point $\left(2,–1,6\right)$ and the equation of plane $3x–2y+z=6$

Consider the point $\left(x_1,y_1,z_1\right)$ and the plane $ax+by+cz=0$ . The distance between the point $\left(x_1,y_1,z_1\right)$ and the plane $ax+by+cz=0$ is

  $D=\frac{|\overline{PQ}\cdot n|}{\parallel n\parallel }$

Where n is obtained by the direction ratios of the plane that is $n=\langle a,b,c\rangle$ and the point Q is given and by letting the points P in the yz-plane, the vector $\overline{PQ}$ is calculated.

Consider the point $\left(2,–1,6\right)$ and the plane $3x–2y+z=6$

Here $n=\langle 3,-2,1\rangle$ is normal to the given plane. It assumes that $Q\left(2,–1,6\right)$

To find a point in the plane, let $y=0$ and $z=0$ andobtain the point $P\left(1,0,0\right).$

The vector form from P to Q is

  $\begin{array}{cc}\overline{PQ}& =\langle 2-1,-1-0,6-0\rangle \\ & =\langle 1,-1,6\rangle \end{array}$

Therefore the distance between the points and the plane is

  $\begin{array}{cc}& D=\frac{|\overline{PQ}\cdot n|}{\parallel n\parallel }\\ & =\frac{|(1)(3)+(-1)(-2)+6(1)|}{\sqrt{9+4+1}}\\ & =\frac{\left|11\right|}{\sqrt{14}}\\ & =\frac{11\sqrt{14}}{14}\end{array}$

Hence, the distance between the points $\left(2,–1,6\right)$ and the plane $3x–2y+z=6$ is $\frac{11\sqrt{14}}{14}$ .